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What is the derivative of #x^(2/3)+y^(2/3)=5# at the given point of #(8,1)#?

Answer 1

Emily Ammerman

#dy/dx = -1/2# at #(x,y) = (8, 1)#

First, let's find #dy/dx# using implicit differentiation:

#d/dx(x^(2/3)+y^(2/3))=d/dx5#

#=> 2/3x^(-1/3) +2/3y^(-1/3)dy/dx = 0#

#=> 2/3y^(-1/3)dy/dx = -2/3x^(-1/3)#

#=> dy/dx = -(x/y)^(-1/3)#

Now, we evaluate #dy/dx# at our given point of #(x,y) = (8,1)#

#dy/dx|_((x,y)=(8,1)) = -(8/1)^(-1/3)#

#=-8^(-1/3)#

#=-1/2#

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Answer 2

Charles Arocha

To find the derivative of (x^{2/3} + y^{2/3} = 5) at the point ((8, 1)), we first implicitly differentiate the equation with respect to (x), and then we substitute the given point's coordinates into the resulting expression to find the derivative's value at that point.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.